$\begin {array} {1 1} (A)\;tautology & \quad (B)\;fallacy \\ (C)\;cannot \: be\: determined & \quad (D)\;None\: of\: these \end {array}$

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Ans : (A)

Assume that $ a \geq b \geq c$. We must have $b+c > a.$

Also note that $b+c \leq c+a \leq a+b$

$ \large\frac{1}{(b+c)} \geq \large\frac{ 1}{(c+a)} \geq \large\frac{1}{(a+b)}$

To show that $ \large\frac{1}{(b+c)},\large\frac{1}{(c+a)},\large\frac{1}{(a+b)}$ are sides of a triangle,

it is sufficient to show that

$ \large\frac{1}{(c+a)} + \large\frac{1}{(a+b)} > \large\frac{1}{(b+c)}$……………(i)

As $a \geq b \geq c$, we get

$2a \geq a+b\: and\: 2a \geq a+c$

$\large\frac{1}{2a} \leq \large\frac{1}{(a+b)}, \large\frac{1}{2a} \leq \large\frac{1}{(a+c)}$

$\large\frac{1}{(a+b)} + \large\frac{1}{(a+c)} \geq \large\frac{1}{2a} + \large\frac{1}{2a} = \large\frac{1}{a} > \large\frac{1}{(b+c)}$ [From Eq. (i)]

So, it represents a triangle.

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